amplifieranalogfeedbackoperational-amplifier
In the above inverting amplifier, we assume -ve node to be at 0V approx (virtual ground), because the open loop gain Aol is very high and hence the differential voltage (Vd = V+ – V- = Vsat/Aol) has to be 0V approx. But why this concept is not applied in the below schmitt trigger ? Should not the differential voltage Vd be = Vsat/Aol here too, despite the positive feedback ?
Your hyperphysics link is correct, and so is your conclusion about the input impedance of the voltage follower. The math follows from the basic control system diagram. You can see a nice presentation on it here:
http://slideplayer.com/slide/1496732/
If you do that the amplifier will amplifiy its own output and the ouput will go high or low and stick there until a sufficiently low or high input is presented, which will make it go the other way (low or high).
In essence you will have created a schmitt trigger.
What the formulas don't show is that the solution for this amplifier configuration describes the metastable point,
Another limitation of those equations is that they don't represent the limited output voltage range of real amplifiers.
Best Answer
The virtual ground principle applies to the classical inverting opamp circuit (with NEGATIVE DC feedback) only. Such a negative DC feedback ensures a fixed DC operating point around which the amplified signal can swing.
For dual DC voltage supply this operational point is identical to a very small DC voltage difference directly across the opamp terminals (µV range). Assuming an IDEAL opamp (infinite DC gain)this voltage difference approaches zero (virtual ground principle).
Such an operation (amplification of an applied input signal around a fixed DC operational point ) requires linear operation of the whole amplifier.
In contrast, an opamp with positive feedback also has a fixed DC output voltage (saturation at app. one of the power rails), which, however, is not called "operational point". In this case, the opamp is not biased for linear operation and cannot be regarded as a linear amplifier with a very large voltafe gain. For this reason, the voltage at the inverting terminal is not very small resp. zero. Instead, this voltage is simply Vout,dc*R1/(R1+R2).
MITURAJ - does this answer the question?